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Question

Determine the value of $r$ for which the integral $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is convergent and evaluate the integral for that value of $r$.

My working

It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0\ $.

If $r \neq 0\ $, then

\begin{align} \int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & = \lim\limits_{a\to\infty}[r\ln(x + 1) - \frac 3 4 \ln|2x^2 + r|]^{x = a}_{x = 1} \\[5 mm] & = \lim\limits_{a\to\infty}[r\ln(a + 1) - \frac 3 4 \ln|2a^2 + r| - r\ln 2 + \frac 3 4 \ln|2 + r|] \\[5 mm] & = \lim\limits_{a\to\infty}\left(r\ln\frac {a + 1} 2 + \frac 3 4 \ln\left|\frac {2 + r} {2a^2 + r}\right|\right) \end{align}


This question just appeared in my final calculus examination this afternoon and I got stuck here ):. We have practiced such questions before but I cannot seem to get this one. I would usually proceed by trying L'Hopital's Rule since we want the limit to exist. I know I probably will not be touching calculus for quite a while from now, but I still want to figure this out. Any intuitions as to how I should proceed will be greatly appreciated :)

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You don't have to care about the exact value of the integral. Simply use asymptotics.

To see how the integrand functions behaves asymptotically as $x \to \infty$, compute the sum of the two fractions.

$$\frac{r}{x+1} - \frac{3x}{2x^2+r} = \frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)}$$

Now consider two cases:

  1. If $2r-3 = 0$, then asymptotically as $x \to \infty$ $$\frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)} \sim \frac{-3x}{2x^3} = O (1/x^2)$$ hence the integral is convergent.

  2. If $2r-3 \neq 0$, then asymptotically as $x \to \infty$ $$\frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)} \sim \frac{(2r-3)x^2}{2x^3} = C \cdot \frac{1}{x}$$ (where $C \neq 0$ is a constant), hence the integral is divergent.

This means that the integral converges only when $r=3/2$.

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By using the comparison test, we know that if we integrate a rational function $\int_a^{\infty} p(x)/q(x)dx$ such that $q(x)\neq 0$ on the interval, then we converge if and only if $\deg(p)-\deg(q)<-1$. However, actually combining and simplifying the rational functions in this problem is tedious algebra, so one can ask if there is a better way. In fact, there is.

Let us write $r/x+1=r/x+a(x)$ where $a(x)$ is some rational function of degree less than $-1$, and let us write $3x/(2x^2+r)=(3/2)/x+b(x)$ where $b(x)$ has degree less than $-1$. We see that if $r=3/2$, we are left with the integral of $a(x)-b(x)$, which will converge. However, for any other choice of $r$, we do not get perfect cancelation and we will be left with something of degree $-1$, so our integral will be divergent.

This allows us to avoid having to integrate anything, or even having to do algebra, simply to compare our integral to the right thing.

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  • $\begingroup$ May I know where the identity/rule "$deg(p) - deg(q) < -1$" comes from? I mean, is there a formal name for it? Or at least, what is the intuition behind it? I ask because I have never heard of it and it was not taught in my calculus class. $\endgroup$ – Ethan Mark Nov 28 '20 at 15:12
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    $\begingroup$ It is a combination of the p-test and comparison test. $\endgroup$ – Aaron Nov 28 '20 at 15:30
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Try to bring the logarithms together! :)

$$\lim_{a\to \infty} \biggl(r\ln\frac{a+1}{2}+\frac 3 4\ln|\frac{2+r}{2a^2+r}|\biggl)$$ You maybe should take $r < -2$ and $r > -2$ because of the absolute value, but it gives the same result so I'll go with normal brackets. $$\lim_{a\to \infty} (\ln(\frac{a+1}{2})^r+ \ln(\frac{2+r}{2a^2+r})^\frac 3 4 )$$ $$\lim_{a\to \infty} \biggl(\ln\bigl((\frac{a+1}{2})^r \cdot (\frac{2+r}{2a^2+r})^\frac 3 4\bigl )\biggl)$$

$$\lim_{a\to \infty} \biggl(\ln\bigl((\frac{(2+r)^\frac 3 4}{2^r}) \cdot (\frac{(a+1)^r}{(2a^2+r)^\frac 3 4})\bigl )\biggl)$$ Now you can split the logarithms if you want, take the minus with the left part when you do $r < -2$. The first part is a constant. The second part is the intertesting one. The limit of the logarithm of that part has to exist. Both numerator and denominator are polynomials of $a$.

If the degree of the top polynomial ist bigger, the fraction will go to infinity and so does the logarithm of it. The integral diverges.

If the degree of the bottom polynomial is bigger, the fraction will go to $0$ and the logarithm to negative infinity. The integral diverges.

Only if both polynomials have the same degree, the fraction will go to a finite number and so does the logarithm. The integral converges. Comparing the degrees gives: $$r = 2*3/4$$ So $r=1.5$. Getting the value of the integral being $0.75\ln(7)-3\ln(2)$ should not be a problem.

Its my first answer on math.stackexchange so my formulas look pretty bad and maybe my answer isnt perfect but I hope I could help you! :P

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I have finally managed to answer the question :)


It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0$.

If $r \neq 0$, note that we can always choose $a$ large enough such that $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x = \int^a_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x + \int^{\infty}_a (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x,$$ where $$\int^a_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is always convergent $\forall\ r \in \mathbb {R}$.

Then, the problem reduces to checking the convergence of $$\int^{\infty}_a (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x,$$ where

\begin{align} \int^{\infty}_a (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & = \lim\limits_{b\to\infty}[r\ln(x + 1) - \frac 3 4 \ln(2x^2 + r)]^{x = b}_{x = a} \\[5 mm] & = \lim\limits_{b\to\infty}[r\ln(b + 1) - \frac 3 4 \ln(2b^2 + r) - r\ln (a + 1) + \frac 3 4 \ln(2a^2 + r)] \\[5 mm] & = \ln \left[\lim\limits_{b\to\infty}\frac {(b + 1)^r} {(2b^2 + r)^{\frac 3 4}}\right] - r\ln (a + 1) + \frac 3 4 \ln(2a^2 + r) \\[5 mm] & = \ln \left[\lim\limits_{b\to\infty}\frac {b^{r - \frac 3 2}(1 + \frac 1 b)^r} {(2 + \frac r {b^2})^{\frac 3 4}}\right] - r\ln (a + 1) + \frac 3 4 \ln(2a^2 + r). \end{align}

Observe that, due to the condition imposed on $a$, the last two terms, namely, $- r\ln (a + 1)$ and $\frac 3 4 \ln (2a^2 + r)$ always exist. Moreover, as $b \rightarrow \infty$, $(1 + \frac 1 b)^r$ and $(2 + \frac r {b^2})^{\frac 3 4}$ are always convergent, so it suffices to check the convergence of $\lim\limits_{b\to\infty}b^{r - \frac 3 2}$ and it is easy to see that $r$ can only take on one value, that is $\frac 3 2$.

Thus, when $r = \frac 3 2$,

\begin{align} \int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & = \lim\limits_{c\to\infty}[\frac 3 2\ln(x + 1) - \frac 3 4 \ln(2x^2 + \frac 3 2)]^{x = c}_{x = 1} \\[5 mm] & = \lim\limits_{c\to\infty}[\frac 3 2\ln(c + 1) - \frac 3 4 \ln\frac {4c^2 + 3} 2 - \frac 3 2\ln 2 + \frac 3 4 \ln\frac 7 2] \\[5 mm] & = \frac 3 4\lim\limits_{c\to\infty}[\ln(c + 1)^2 - \ln(4c^2 + 3)] - \frac 3 2\ln 2 + \frac 3 4 \ln 7 \\[5 mm] & = \frac 3 4\lim\limits_{c\to\infty}\ln\frac {(c + 1)^2} {4c^2 + 3} - \frac 3 2\ln 2 + \frac 3 4 \ln 7 \\[5 mm] & = \frac 3 4\ln\frac 1 4 - \frac 3 2\ln 2 + \frac 3 4 \ln 7 \\[5 mm] & = \frac 3 4 \ln 7 - 3\ln 2 \end{align}

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